Question: Let $f$ be a twice differentiable function, and let $f(-3)=-4$, $f'(-3)=0$, and $f''(-3)=1$. What occurs in the graph of $f$ at the point $(-3,-4)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-3,-4)$ is a minimum point. (Choice B) B $(-3,-4)$ is a maximum point. (Choice C) C There's not enough information to tell.
Answer: Since $f'(-3)=0$, we know that $x=-3$ is a critical point. The second derivative test allows us to analyze what happens in the graph of $f$ at this point according to these three cases: If $f''(-3)>0$, the graph of $f$ has a minimum point at $x=-3$. If $f''(-3)<0$, the graph of $f$ has a maximum point at $x=-3$. If $f''(-3)=0$, the test is inconclusive. [Why is this so?] We are given that $f''(-3)=1>0$. Therefore, $(-3,-4)$ is a minimum point.